Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 34D (General Relativity)

Write $f=1-r^2/a^2$ and $r_{\ast }=a\mathrm{artanh}(r/a)$. Since $d{r}_{\ast }/dr=f^{-1}$, $dt=du+dr/f$. Substitution gives the outgoing Outgoing Eddington-Finkelstein coordinatesThe radial block has determinant $-1$, independent of $f$, and all coefficients are smooth at $r=a$. This is a coordinate singularity of the static chart, not a spacetime singularity. The transformation was defined inside the horizon, but the new metric itself extends smoothly across it.

At $r=0$ the spherical angular coordinates degenerate as usual. Cartesian spatial coordinates make the original static metric regular there: $f=1+O(r^2)$ and the extra spatial radial correction is $x_i{x}_j/[a^2(1-r^2/a^2)]$. Thus the origin is not a spacetime singularity either.

The inverse radial block is $\left(\begin{array}{cc}0& -1\\ -1& f\end{array}\right)$. Since $u_{,a}$ has only its $u$ component nonzero, $g^{ab}{u}_{,a}{u}_{,b}=g^{uu}=0$. Therefore $\boxed{\nabla u=-{\partial }_r\text{ is null}}$; its integral curves lie on the outgoing null hypersurfaces of constant $u$.

Radial null curves obey $-fd{u}^2-2dudr=0$. For their unparameterized paths,Inside the horizon, the first family satisfies $dt/dr=1/f$, so $dr/dt=f>0$ and is outgoing. The second satisfies $dt/dr=-1/f$ and is ingoing. Integrating gives $u=u_0$ for outgoing rays and $u=C-2a\mathrm{artanh}(r/a)$ for ingoing rays. The null paths are geodesics after suitable reparameterization; for example in the two-dimensional radial metric each null direction has covariant acceleration proportional to itself.

To show constant-$u$ rays at the requested angle, the diagram uses vertical plotting coordinate $u+r$ and horizontal $r$, with equal scale. This is the stated tilted $u$ grid: constant $u$ becomes a line of slope one. The ingoing curves bend to decreasing $u+r$ as $r$ approaches the horizon.

For a radial timelike geodesic, the static Killing energy is $E=f\dot{t}$ and normalization gives ${\dot{r}}^2=E^2-f$. Choose $E=\sqrt{2}$ and start at $r(0)=0$ with outward velocity one. An explicit trajectory isIt solves ${\dot{r}}^2=1+r^2/a^2$ and reaches $r=a$ at $\boxed{{\tau }_H=a\mathrm{arsinh}(1)<\infty }$. Differentiating the energy equation gives $\ddot{r}=r/a^2$, the radial geodesic equation, so this is an actual freely falling trajectory.

The regular coordinate satisfies $\dot{u}=(E-\dot{r})/f=1/(E+\dot{r})$, which is finite and positive at the crossing. Together with $u(\tau )=u(0)+{\int }_0^{\tau }[\sqrt{2}+\cosh (s/a){]}^{-1}ds$, the trajectory extends through the horizon in the regular chart. Divergence of the static time there does not imply infinite proper time.